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4 years ago

How can the Distributive Property help you rewrite expressions ?

Mathematics

2 answers:

givi [52]4 years ago

8 0

The distributive property helps re-write expressions because it simplifies expressions. <span />

Feliz [49]4 years ago

6 0

For example, instead of directly solving 5(2+5), you can use the distributive property to turn that expression into 5(2) + 5(5). This equates to 10 + 25 = 35

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How do you do this math 18-0.4<br> in long subtraction

Gwar [14]

18.0 1
- 0.4 17.0
_____ - 0.4
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17.6

5 0

4 years ago

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Simplify the expressions by combining like terms<br><br> 4r + 9r - 11r + 7r

konstantin123 [22]

9r
Explanation:
4r + 9r = 13r
13r-11r = 2r
2r + 7r = 9r

4 0

3 years ago

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Prove the divisibility of the following numbers:

Brut [27]

Answer:

Step-by-step explanation:

To prove divisibility, we need to factor the divident such that one of its factors matches the divisor.

(I use the notation x|y to denote that x divides y)

(A)

$75^{30}|45^{45}\cdot15^{15}\\45^{45}\cdot15^{15}=3^{45}\cdot 15^{45}\cdot 15^{15}=\\=3^{45}\cdot 15^{60}=3^{45}\cdot 15^{30}\cdot 15^{30}=3^{45}\cdot (3\cdot5)^{30}\cdot 15^{30}=\\=3^{45}\cdot 3^{30}\cdot(5\cdot 15)^{30}=3^{45}\cdot 3^{30}\cdot(75)^{30}\\\implies\\75^{30}|3^{45}\cdot 3^{30}\cdot75^{30}$

(B)

$72^{63}|24^{54}\cdot 54^{24}\cdot2^{10}\\24^{54}\cdot 54^{24}\cdot2^{10}=(2^{162}\cdot 3^{54})\cdot(2^{24}\cdot 3^{72}) \cdot 2^{10}\\=2^{196}\cdot 3^{126}$

In this case, it is easier to also factor the divisor to primes:

$72^{63}=2^{189}\cdot 3^{126}$

Both of these factor must be matched in the dividend in order to prove divisibility, and that indeed turns out to be true:

$2^{189}\cdot 3^{126}|2^{196}\cdot 3^{126}\implies\\2^{189}|2^{196}\,\,\mbox{and}\,\,3^{126}|3^{126}$

6 0

3 years ago

Please help me with this question!!!!!

Naddika [18.5K]

Answer:

Equation of line in slope-intercept form that passes through (4, -8) and is perpendicular to the graph $y= \frac {2}{5} x - 3$ is below

$y = - \frac {5}{2} x + 2$

Step-by-step explanation:

Slope of the equation $y= \frac{2}{5} x -3$ is $m_1 = \frac{2}{5}$

Since slopes of perpendicular lines are negative reciprocal of each other, therefore slope of other line is given as

$m_2 = - \frac {1}{m_1} = - \frac {5}{2}$

Equation of line in point slope form is given as

$y-y_1=m_2(x-x_1)$

Here (x1, y1) = (4, -8)

$y+8 = - \frac{5}{2} (x-4)$

Simplifying it further

$y = - \frac {5}{2} x + \frac {5}{2} 4 - 8$

$y = - \frac {5}{2} x + 2$

6 0

3 years ago

A manufacturing company produces steel housings for electrical equipment. The main component part of the housing is a steel trou

Mrrafil [7]

Answer:

Check the explanation

Step-by-step explanation:

(a)

H0: Population mean = 8.46

H1: Population mean is not equal to 8.46

The test statistic (t) is given by the following expression -

${t=\frac{\overline{x}-\mu_0}{s/\sqrt{n}}}$

We have calculated the sample mean (8.421) and sample standard deviation (0.0461). Therefore, the test statistic (t) will be -

t = (8.421 - 8.46)/(0.0461/sqrt(49)) = -5.92

The left-sided critical value from t-distribution (two-tailed) is -2.01. Thus the test statistic falls in the rejection region and we reject the null hypothesis at 95% confidence level and conclude that the population mean is different from 8.46 inches.

(b)

There are the following four assumptions for a single sample t-test.

The observations are in ratio scale.

The observations have been taken in such a manner that every single observation is independent and uncorrelated from the others.

There should not be any significant outliers in the sample observations.

The sample data should be approximately normal.

(c)

The first assumption is true as the data is in inches. The second assumption cannot be tested now. It will depend upon the situation and study design which we assume as correct in this case. The third assumption is checked by plotting a box plot and finding the outliers. The final assumption can be checked using the Anderson-Darling normality test.

Kindly check the graphical table in the attached images below.

(d)

The data is normal (the p-value of Anderson-Darling test in > 0.05). However, two points in the Box plot are outliers though not grossly different from the entire sample data set. So, we conclude that the assumptions are valid in this case for conducting the t-test.

7 0

4 years ago